chemistry

When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)

please explain realy realy well

thanks:)

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  1. Here is what I posted previously.

    Convert 12 ounces to grams.
    (mass ice x 330 J/g) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0
    Solve for mass ice.

    Show your work and I'll try to find the error. Mass ice is the only unknown. mass water is 12 ounces; you need to convert that to grams. Specific heat water you know as well as Tf and Ti. There is no explaining to do.

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  2. this is wut i got:

    12oz=340.194 grams

    (miX330j/g) + [(340.194gX4.184X40]=0

    56934.86 + 330j/g(mi)=0

    -56934.86=330j/g(mi)

    -56934.86/330j/g = -172.5298=(mi)

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  3. oops sorry I am rebekah

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  4. my sis changed it 4 her

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  5. ok. The error is in the 40. It should be a -40. It works this way.
    (mass ice x 330) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0

    (330X) + [(340.2 x 4.184 x (35-75)] = 0
    330X + [(340.2 x 4.184 x (-40)] = 0
    I will let you finish but I think you can see that it changes only the sign so you end up with +172 g and not -172 g(the negative sign makes no sense, of course).

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  6. Some quick comments. The F must be converted to C, the ice, when it melts at zero C must also be heated to 35 F (converted to C), and the fluid ounce is not the same as the the ounce in mass units.

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  7. how much ice is required to cool a 10.0 oz drink from 80 f to 31 f, if the heat capacity of the drink is 4.18 j/gc?

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  8. This answer is fine because -40 F is equal to -40 C
    I got the answer 172 g of ice needed.

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